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Math Help - Integration by Parts

  1. #1
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    Integration by Parts

     \int{x^2tan^{-1}xdx}

    \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx

    let {}u=1+x^2, \frac{du}{2}=xdx

    \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)

    \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)

    In the answer, there is no -1/6. Help?
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  2. #2
    Behold, the power of SARDINES!
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    Quote Originally Posted by quantoembryo View Post
     \int{x^2tan^{-1}xdx}

    \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx

    let {}u=1+x^2, \frac{du}{2}=xdx

    \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int (1-1/u)

    \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)

    In the answer, there is no -1/6. Help?
    Since you have an indefinate itegral you need an arbitarty constnat +C the -\frac{1}{6} can be combined into the constant.
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  3. #3
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    So would you always get rid of terms without a variable?
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  4. #4
    Behold, the power of SARDINES!
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    Quote Originally Posted by quantoembryo View Post
    So would you always get rid of terms without a variable?
    Remember that anti derivatives are only unique up to an arbitrary constant.

    You can check that you have a correct answer by taking the derivative.

    Notice that if you take the derivative of  C or C-\frac{1}{6} both derivatives are zero. Also notice that if you were given limits of integration you would end up with \displaystyle -\frac{1}{6}-\left( -\frac{1}{6}\right)=0

    In my opinion I always combine constants in my C
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  5. #5
    Behold, the power of SARDINES!
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    Also I just noticed I don't see where you got you \frac{1}{6}. I don't think it should be there. Double check you integration done after integrating by parts
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  6. #6
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    Possibly the OP simplified

    \displaystyle{\frac{x^3}{1 + x^2}}

    to

    \displaystyle{1 - \frac{x}{1 + x^2}}

    which should have been

    \displaystyle{\frac{x(1 + x^2) - x}{1 + x^2}}\ =\ \displaystyle{x - \frac{x}{1 + x^2}}

    [Edit: no, that doesn't explain it.]

    Anyway, just in case a picture helps...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight continuous lines differentiate downwards (integrate up) with respect to x. And,



    ... is lazy integration by parts, doing without u and v.



    Further edit: (See below.) Yes, false alarm. Substituting where they did, the OP integrated kind of like this:



    ... so the sixth was just an extra constant as The Empty Set at first thought. Pardon the fuss.

    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
    Last edited by tom@ballooncalculus; February 2nd 2011 at 12:51 PM.
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  7. #7
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    Quote Originally Posted by quantoembryo View Post
     \int{x^2tan^{-1}xdx}

    \int{x^2tan^{-1}xdx} = \frac{x^3}{3}tan^{-1}x-\frac{1}{3}\int \frac {x^3}{1+x^2}dx

    let {}u=1+x^2, \frac{du}{2}=xdx

    \frac{x^3}{3}tan^{-1}x- \frac{1}{6}\int{(1-1/u)}du

    \frac{x^3}{3}tan^{-1}x-(x^2/6)-(1/6)+(1/6)ln(x^2+1)+C

    In the answer, there is no -1/6. Help?
    Your work is fine.

    \displaystyle\frac{2x^3tan^{-1}x-x^2+ln\left[1+x^2\right]}{6}+\left(C-\frac{1}{6}\right)

    and what you have over on the right is a constant, so the fraction can de discarded.
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